The mortality problem for $3\times 3$ matrices: given a finite set $F$ of $3\times 3$
integer matrices, decide whether the zero matrix is a product of members of $F$
(with repetitions allowed). This was proved unsolvable by Michael Paterson,
Studies in Applied Mathematics 49 (1970), 105--107, doi: 10.1002/sapm1970491105.
The corresponding problem for $2\times 2$ matrices is apparently still open.
Edit (11 September 2016): The problem for $2\times 2$ matrices is apparently still open, despite the solution of a seemingly similar problem mentioned in Igor Potapov's answer below.
Thanks for clarifying your question. The formulation that
you and Dorais give seems perfectly reasonable. You have a
first order language for category theory, where you can
quantify over objects and morphisms, you can compose
morphisms appropriately and you can express that a given
object is the initial or terminal object of a given
morphism. In this language, one can describe various finite
diagrams, express whether or not they are commutative, and
so on. In particular, one can express that composition is
associative, etc. and describe what it means to be a
category in this way.
The question now becomes: is this theory decidable? In
other words, is there a computable procedure to determine,
given an assertion in this language, whether it holds in
all categories?
The answer is No.
One way to see this is to show even more: one cannot even
decide whether a given statement is true is true in all
categories having only one object. The reason is that group
theory is not a decidable theory. There is no computable
procedure to determine whether a given statement in the
first order language of group theory is true in all groups.
But the one-point categories naturally include all the
groups (and we can define in a single statement in the
category-theoretic language exactly what it takes for the
collection of morphisms on that object to be a group).
Thus, if we could decide category theory, then we could
decide the translations of the group theory questions into
category theory, and we would be able to decide group
theory, which we can't. Contradiction.
The fundamental obstacle to decidability here, as I
mentioned in my previous answer (see edit history), it the
ability to encode arithmetic. The notion of a strongly
undecidable
structure
is key for proving various theories are undecidable. A
strongly undecidable theory is a finitely axiomatizable
theory, such that any theory consistent with it is
undecidable. Robinson proved that there is a strongly
undecidable theory of arithmetic, known as Robinson's Q. A
strongly undecidable structure is a structure modeling a
strongly undecidable theory. These structures are amazing,
for any theory true in a strongly undecidable structure is
undecidable. For example, the standard model of arithmetic,
which satisfies Q, is strongly undecidable. If A is
strongly undecidable and interpreted in B, then it follows
that B is also strongly undecidable. Thus, we can prove
that graph theory is undecidable, that ring theory is
undecidable and that group theory is undecidable, merely by
finding a graph, a ring or a group in which the natural
numbers is interpreted. Tarski found a strongly undecidable
group, namely, the group G of permutations of the integers
Z. It is strongly undecidable because the natural numbers
can be interpreted in this group. Basically, the number n
is represented by translation-by-n. One can identify the
collection of translations, as exactly those that commute
with s = translation-by-1. Then, one can define addition as
composition (i.e. addition of exponents) and the divides
relation is definable by: i divides j iff anything that
commutes with si also commutes with
sj. And so on.
I claim similarly that there is a strongly undecidable
category. This is almost immediate, since every group can
be viewed as the morphisms of a one-object category, and
the group is interpreted as the morphisms of this category.
Thus, the category interprets the strongly undecidable
group, and so the category is also strongly undecidable. In
particular, any theory true in the category is also
undecidable. So category theory itself is undecidable.
Best Answer
Here is a proof based on the recursion theorem, rather than a reduction of an undecidable problem.
Rice's Theorem. Suppose that $P$ is any set of computable functions, which is not empty and not all computable functions. Then the set $\{ e\mid \varphi_e\in P\}$ is not decidable, where $\varphi_e$ is the function computed by program $e$.
In other words, there is no general procedure to determine from a program whether the function it computes has property $P$ or not.
Proof. Suppose that the set were decidable. Fix a computable function $f$ that is in $P$, and another computable function $g$ that is not in $P$. Now, for any program $e$, let $h(e)$ be the program that on input $n$ first determines whether $\varphi_e\in P$; if so, it outputs $g(n)$, and otherwise $f(n)$. So $\varphi_{h(e)}$ is either $g$ or $f$, depending on whether $\varphi_e\in P$ or not, respectively (but note that we are using the opposite function). In particular, we'll have $$\varphi_e\in P\quad\iff\quad\varphi_{h(e)}\notin P.$$ Meanwhile, by the recursion theorem, there is a program $e$ such that $\varphi_e=\varphi_{h(e)}$, which now gives an immediate contradiction, since $\varphi_e$ and $\varphi_{h(e)}$ are supposed to be opposite with respect to $P$. QED